2020-M2-04

$\begin{array}{cl}
= & \pi \dint_0^1 \left(4x(1-x^2)^\frac{1}{4}\right)^2 dx \\
= & \pi \dint_0^1 \left( 16x^2(1-x^2)^\frac{1}{2}\right) dx
\end{array}$

Let $x=\sin \theta$, then $dx=\cos \theta d\theta$.

When $x=0$, $\theta=0$.

When $x=1$, $\theta=\dfrac{\pi}{2}$.

Hence, the required volume

$\begin{array}{cl}
= & \pi\dint_0^\frac{\pi}{2} 16\sin^2 \theta(1-\sin^2\theta)^\frac{1}{2} \cos\theta d\theta \\
= & 16\pi\dint_0^\frac{\pi}{2} \sin^2\theta\cos^2\theta d\theta \\
= & 4\pi \dint_0^\frac{\pi}{2} (2\sin\theta\cos\theta)^2 d\theta \\
= & 4\pi\dint_0^\frac{\pi}{2} \sin^2 2\theta d\theta \\
= & 2\pi\dint_0^\frac{\pi}{2} \sin^2 2\theta d(2\theta) \\
= & 2\pi \left[\dfrac{(2\theta)}{2}-\dfrac{1}{4}\sin 2(2\theta) \right]_0^\frac{\pi}{2} \text{, by the result of (a).} \\
= & 2\pi \left( \dfrac{\pi}{2} -\dfrac{1}{4}\sin4(\dfrac{\pi}{2})-0+\dfrac{1}{4}\sin 4(0)\right) \\
= & \pi^2
\end{array}$